Proving $364 \mid n^{91} - n^7$ [Generalization of Euler & Fermat Theorems]

Question

I am trying to prove this statement which is equivalent to show $n^{91} = n^7 \pmod{ 364}$. By splitting modulus theorem $n^{91} = n^7 \bmod 91$ and $\bmod 4$. Then I don’t know what to do next... Can anyone help me?

Answer 1

Apply below Euler-Fermat generalization with $\,e = 7,\, f = 84,\ m=\prod p_i^{e_i} = 2^2\cdot 7\cdot 13 $

Theorem $ $ Suppose that $\ m\in \mathbb N\ $ has the prime factorization $\:m = p_1^{e_{1}}\cdots\:p_k^{e_k}\ $ and suppose also that for all $\,i\!:\,$ $\ \color{#0a0}{e_i\le e}\ $ and $\ \color{#c00}{\phi(p_i^{e_{i}})\mid f}.\ $ Then $\ m\mid \color{darkorange}{a^e}(\color{#0af}{a^f\!-1})=:N,\, $ for all $\: a\in \mathbb Z.$

Proof $\ $ For $\,q := p_i^{e_{i}}$ it suffices to prove $\,q\mid \color{darkorange}{a^e}\,$ or $\,q\mid \color{#0af}{a^f\!-1}\,$ since then all $\,p_i^{e_i}\mid N\,$ hence their lcm = product = $\,m\,$ must also divide $N.\,$ If $\ p_{\:\!i}\mid a\ $ then $\,q=p_i^{\color{#0a0}{e_i}}\mid {\color{darkorange}a^{\color{#0a0}e}}\,$ by $\ \color{#0a0}{e_i \le e}.\: $ Else $\,p_i\nmid a\,$ so by $ $ Euler's phi theorem: $\bmod q\!: \ a^{\large \color{#c00}{\phi(q)}}\equiv 1,\:$ hence $\:\color{#0af}{a^{\large \color{#c00}f}\equiv 1},\,$ by $\,\color{#c00}{\phi(q)\mid f}$.

Examples $\ $ There are many instructive examples in prior questions, e.g. below

$\qquad\qquad\quad$ $24\mid a^3(a^2-1)$

$\qquad\qquad\quad$ $40\mid a^3(a^4-1)$

$\qquad\qquad\quad$ $88\mid a^5(a^{20}\!-1)$

$\qquad\qquad\quad$ $6p\mid a\,b^p - b\,a^p$

Answer 2

Use Euler's theorem:

Mod $4$: either $n\equiv0 $ or $2$, so $n^{91}\equiv n^7\equiv0$, or $\color{blue}{n^2\equiv 1}$ so $n^{91}\equiv n^{84}n^7\equiv (n^2)^{42}n^7\equiv n^7$

Mod $7$: either $n\equiv0$, so $n^{91}\equiv n^7\equiv 0$, or $\color{blue}{n^6\equiv 1}$ so $n^{91}\equiv n^{84}n^7\equiv(n^6)^{14}n^7\equiv n^7$

Mod $13$: either $n\equiv0$, so $n^{91}\equiv n^7\equiv 0$, or $\color{blue}{n^{12}\equiv 1}$ so $n^{91}\equiv n^{84}n^7\equiv(n^{12})^{7}n^7\equiv n^7$